blob: b3a9645ecb59473f9bcf0b5e5c432f526ce7c210 (
plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
|
// A Texbook on POWER SYSTEM ENGINEERING
// A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar
// DHANPAT RAI & Co.
// SECOND EDITION
// PART II : TRANSMISSION AND DISTRIBUTION
// CHAPTER 11: LOAD FREQUENCY CONTROL AND LOAD SHARING OF POWER GENERATING SOURCES
// EXAMPLE : 11.10 :
// Page number 336
clear ; clc ; close ; // Clear the work space and console
// Given data
load_1 = 10000.0 // Total balanced load(kW)
V = 33000.0 // Voltage(V)
PF_1 = 0.8 // Lagging power factor
R = 1.6 // Resistance of feeder(ohm/phase)
X = 2.5 // Reactance of feeder(ohm/phase)
load_2 = 4460.0 // Load delivered by feeder(kW)
PF_2 = 0.72 // Lagging power factor
// Calculations
I = load_1*1000/(3**0.5*V*PF_1)*exp(%i*-acos(PF_1)) // Total line current(A)
I_1 = load_2*1000/(3**0.5*V*PF_2)*exp(%i*-acos(PF_2)) // Line current of first feeder(A)
I_2 = I-I_1 // Line current of first feeder(A)
Z_1 = complex(R,X) // Impedance of first feeder(ohm)
Z_2 = I_1*Z_1/I_2 // Impedance of second feeder(ohm)
// Results
disp("PART II - EXAMPLE : 11.10 : SOLUTION :-")
printf("\nImpedance of second feeder, Z_2 = %.2f∠%.1f° ohm \n", abs(Z_2),phasemag(Z_2))
printf("\nNOTE: ERROR: Changes in the obtained answer from that of textbook is due to wrong values of substitution")
|
